diff --git a/decode-ways/KwonNayeon.py b/decode-ways/KwonNayeon.py
new file mode 100644
index 000000000..f2f523bbb
--- /dev/null
+++ b/decode-ways/KwonNayeon.py
@@ -0,0 +1,42 @@
+"""
+Conditions:
+- 1 <= s.length <= 100
+- s contains only digits and may contain leading zero(s).
+
+Time Complexity: O(n)
+- 각 문자를 한 번씩만 방문함 (문자열의 길이: n)
+
+Space Complexity: O(n)
+- dp 배열의 크기는 n+1
+- 이 배열 이외에 추가 공간을 사용하지 않음
+
+Base cases:
+- If string is empty or starts with '0', return 0 (impossible to decode)
+
+Dynamic Programming approach:
+- dp[i] represents the number of ways to decode first i characters
+- dp array has size n+1 to include the empty string case (dp[0])
+- For each position, consider both single-digit and two-digit decodings
+
+메모:
+- 다른 dp문제 풀어보기
+"""
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        if not s or s[0] == '0':
+            return 0
+
+        n = len(s)
+        dp = [0] * (n + 1)
+        dp[0] = 1
+
+        for i in range(1, n + 1):
+            if s[i-1] != '0':
+                dp[i] += dp[i-1]
+
+            if i > 1 and s[i-2] != '0':
+                two_digit = int(s[i-2:i])
+                if 1 <= two_digit <= 26:
+                    dp[i] += dp[i-2]
+
+        return dp[n]
diff --git a/maximum-subarray/KwonNayeon.py b/maximum-subarray/KwonNayeon.py
new file mode 100644
index 000000000..d12a244b9
--- /dev/null
+++ b/maximum-subarray/KwonNayeon.py
@@ -0,0 +1,31 @@
+"""
+Conditions:
+- 1 <= nums.length <= 10^5
+- -10^4 <= nums[i] <= 10^4
+
+Time Complexity: O(n)
+- 배열을 한 번만 순회함
+
+Space Complexity: O(1)
+- 두 변수 이외에 추가 공간을 사용하지 않음
+
+풀이방법: 
+1. Base case: If nums is empty, return 0
+2. Initialize variables (current_sum and max_sum as the first value in the array)
+3. Traverse from the value at 1st index to the last, update current_sum
+   - Decide whether to add the current value (num) to the existing subarray or start a new one
+4. Update max_sum
+   - Choose the larger between the updated current_sum and the previous max_sum
+"""
+class Solution:
+    def maxSubArray(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+
+        current_sum = max_sum = nums[0]
+
+        for num in nums[1:]:
+            current_sum = max(num, current_sum + num)
+            max_sum = max(current_sum, max_sum)
+
+        return max_sum
diff --git a/valid-palindrome/KwonNayeon.py b/valid-palindrome/KwonNayeon.py
index 5bb039da8..c8a7cc970 100644
--- a/valid-palindrome/KwonNayeon.py
+++ b/valid-palindrome/KwonNayeon.py
@@ -1,21 +1,12 @@
 """
-Title: 215. Valid Palindrome  
-Link: https://leetcode.com/problems/valid-palindrome/  
-
-Summary:
-    - Palindrome이라면 True, 아니라면 False를 반환하는 문제.
-    - Palindrome이란, 앞으로 읽어도 뒤에서부터 읽어도 동일한 단어를 뜻함.
-    - 추가 조건: 대소문자를 구분하지 않으며, 알파벳과 숫자 이외의 문자는 제거해야 함.
-    - e.g. racecar
-
 Conditions:
-    - 입력 문자열이 Palindrome인 경우: `True` 반환
-    - Palindrome이 아닌 경우: `False` 반환
+- 1 <= s.length <= 2 * 10^5
+- s consists only of printable ASCII characters.
+
+<Solution 1>
+Time Complexity: O(n)
 
-Time Complexity:
-    - O(n)
-Space Complexity:
-    - O(n)
+Space Complexity: O(n)
 """
 class Solution:
     def isPalindrome(self, s: str) -> bool:
@@ -23,4 +14,29 @@ def isPalindrome(self, s: str) -> bool:
         if s == s[::-1]:
             return True
         return False
+"""
+<Solution 2>
+Time Complexity: O(n) 
+- 팰린드롬일 경우, 각 문자를 한 번씩 검사
+
+Space Complexity: O(1)
+- left, right 포인터 외에 추가 공간 사용 없음
+"""
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        left, right = 0, len(s) - 1
+
+        while left < right:
+            while left < right and not s[left].isalnum():
+                left += 1
+
+            while left < right and not s[right].isalnum():
+                right -= 1
+
+            if s[left].lower() != s[right].lower():
+                return False
+
+            left += 1
+            right -= 1
 
+        return True